ee 340 power transformers -...

EE 740

Power Transformers

Y. Baghzouz

Spring 2013

Preliminary considerations

A transformer is a device that converts one AC voltage to

another AC voltage at the same frequency. It consists of one

or more coil(s) of wire wrapped around a common

ferromagnetic core.

• In practice, the windings are wrapped on top of

each other to minimize flux leakage.

Ideal transformer

Consider a lossless transformer with an input (primary) winding

having Np turns and an output (secondary) winding of Ns turns.

The relationship between the voltage applied to the primary winding

vp(t) and the voltage produced on the secondary winding vs(t) is

( )

( )

p p

s s

v t Na

v t N

where a is the turn ratio of the transformer.

An ideal transformer (unlike the real one) can be characterized as

follows:

1.The core has no hysteresis nor eddy currents.

2.The magnetization curve is vertical with no saturation

3.The leakage flux in the core is zero.

4.The resistance of the windings is zero.

Ideal transformer

The relationship between the primary ip(t) and secondary is(t) currents is

( ) 1

( )

p

s

i t

i t a

Phasor notation:

p

s

aV

V

1p

s a

I

I

• The phase angles of primary and secondary voltages are the same.

• The phase angles of primary and secondary currents are the same also.

• The ideal transformer changes magnitudes of voltages and currents but not

their angles.

Ideal Transformer

• One winding’s terminal is usually marked by a dot used to

determine the polarity of voltages and currents.

• If the voltage is positive at the dotted end of the primary

winding at some moment of time, the voltage at the dotted end

of the secondary winding will also be positive at the same time

instance.

• If the primary current flows into the dotted end of the primary

winding, the secondary current will flow out of the dotted end

of the secondary winding.

Power in an ideal transformer

Assuming that p and s are the angles between voltages and currents on the

primary and secondary windings respectively, the power supplied to the

transformer by the primary circuit is:

cosin p p pP V I

The power supplied to the output circuit is

cosout s s sP V I

Since ideal transformers do not affect angles between voltages and currents:

p s

Power in an ideal transformer

Since for an ideal transformer the following holds:

;p

s s p

VV I aI

a

Therefore:

The output power of an ideal transformer equals to its input power – to be

expected since assumed no loss. Similarly, for reactive and apparent powers:

sin sinout s s p p inQ V I V I Q

out s s p p inS V I V I S

cos cos cosout i

p

s s p p p n

VV I aIP V I

aP

Impedance transformation

The impedance is defined as a following ratio of phasors:

L L LZ V I

A transformer changes voltages and currents and, therefore, an apparent

impedance of the load that is given by

L s sZ V I

The apparent impedance of the

primary circuit is:

'L p pZ V I

which is

2 2'p s s

p s s

L L

aa

aZ a Z

V V V

I I I

Analysis of circuits containing ideal transformers: Example

Example 4.1: a) What is the voltage at the load? Calculate the transmission line

losses? b) If a 1:10 step up transformer and a 10:1 step down transformer are

placed at the generator and the load ends of the transmission line respectively,

what are the new load voltage and the new transmission line losses?

a) Without transformers:

480 0

0.18 0

90.8 37

.24 4 3

480 0

5.29 37.8

.8

G line load

line load

A

j j

VI I I

Z Z

90.8 37.8 (4 3) 90.8 37.8 5 36.9 454 0.9load load load j V V I Z

2 290.8 0.18 1484loss line lineP I R W


b) With transformers, we will • eliminate transformer T2

by referring the load over to the transmission line’s voltage level.

• Eliminate transformer T1 by referring the transmission line’s voltage level to the source side,

'

480 0

5.003 36.88

95.94 36.88

G

eq

A

VI

Z


Knowing transformers’ turn ratios, we can determine line and load currents:

1 0.1 95.94 36.88 9.594 36.88line Ga A I I

2 10 9.594 36.88 95.94 36.88load linea A I I

Therefore, the load voltage is:

95.94 36.88 5 36.87 479.7 0.01load load load V V I Z

The losses in the line are:

2 29.594 0.18 16.7loss line lineP I R W

Real transformer

A portion of the flux produced in

the primary coil passes through

the secondary coil (mutual flux);

the rest passes through the

external medium (leakage flux):

p m Lp

leakage primary flux mutual flux

Similarly, for the secondary coil:

s m Ls

Leakage secondary flux

Flux leakage

Real transformer

From the Faraday’s law, the primary coil’s voltage is:

( ) ( ) ( )p Lpm

p p p p p Lp

d ddv t N N N e t e t

dt dt dt

The secondary coil’s voltage is:

( ) ( ) ( )s m Lss s s s s Ls

d d dv t N N N e t e t

dt dt dt

The primary and secondary voltages due to the mutual flux are:

( ) mp p

de t N

dt

( ) Ls

s s

de t N

dt

Combining the last two equations:

( ) ( )p m s

p s

e t d e t

N dt N

Real transformer

Therefore:

( )

( )

p p

s s

e t Na

e t N

That is, the ratio of the primary voltage to the secondary voltage both caused by

the mutual flux is equal to the turns ratio of the transformer.

The following approximation normally holds since the leakage flux is much

smaller than the mutual flux;:

( )

( )

p p

s s

v t Na

v t N

The magnetization current in a real transformer

Even when no load is connected to the secondary coil of the transformer, a

current will flow in the primary coil. This current consists of:

1. The magnetization current im is needed to produce the flux in the core;

2. The core-loss current ih+e corresponds to hysteresis and eddy current

losses.

Flux causing the

magnetization current

Typical magnetization curve

Excitation Current in a real transformer

Core-loss current

Core-loss current is:

total excitation current in a transformer

1. Nonlinear due to nonlinear effects of hysteresis;

2. In phase with the voltage.

The total no-load current in the core is called the excitation current of

the transformer:

ex m h ei i i

The current ratio on a transformer

If a load is connected to the secondary coil, there will be a current flowing

through it.

A current flowing into the dotted

end of a winding produces a

positive magnetomotive force F:

p p pF N is s sF N i

The net magnetomotive force in the

core is

net p p s sF N i N i

For well-designed transformer cores, the reluctance is very small if the

core is not saturated. Therefore:

0net p p s sF N i N i 1p

s

ps

p s

p

s

i N

i N aN i N i

The transformer’s equivalent circuit

To model a real transformer accurately, we need to account for

the following losses:

1. Copper losses – resistive heating in the windings: I2R.

2. Eddy current losses – resistive heating in the core:

proportional to the square of voltage applied to the

transformer.

3. Hysteresis losses – energy needed to rearrange magnetic

domains in the core: nonlinear function of the voltage

applied to the transformer.

4. Leakage flux – flux that escapes from the core and flux that

passes through one winding only.

The exact equivalent circuit of a real transformer

• Cooper losses are modeled by the resistors Rp and Rs.

• The leakage flux can be modeled by primary and secondary inductors.

• The magnetization current can be modeled by a reactance XM

connected across the primary voltage source.

• The core-loss current can be modeled by a resistance RC connected

across the primary voltage source.

• Both magnetizing and core loss currents are nonlinear; therefore, XM

and RC are just approximations.

The exact equivalent circuit of a real transformer

The equivalent circuit is usually referred to the primary side

or the secondary side of the transformer.

Equivalent circuit of the

transformer referred to its

primary side.

Equivalent circuit of the

transformer referred to its

secondary side.

Approximate equivalent circuit of a transformer

Without an excitation branch

referred to the primary side.

Referred to the secondary side.

Without an excitation branch

referred to the secondary side.

Referred to the primary side.

Determining the values of components

The open-circuit test.

Full line voltage is applied to the primary

side of the transformer. The input voltage,

current, and power are measured.

From this information, the power factor of the input current and the magnitude and

the angle of the excitation impedance can be determined.

To evaluate RC and XM, we define the conductance of the core-loss resistance and

The susceptance of the magnetizing inductor :

1C

C

GR

1

M

M

BX


1 1E C M

C M

Y G jB jR X

ocE

oc

IY

V

cos oc

oc oc

PPF

V I

Since both elements are in parallel, their admittances add. Therefore, the total

excitation admittance is:

The magnitude of the excitation admittance in the open-circuit test is:

The angle of the admittance in the open-circuit test can be found from the circuit

power factor (PF):


1cosoc ocE

oc oc

I IY PF

V V

In real transformers, the power factor is always lagging, so the angle of the current

always lags the angle of the voltage by degrees. The admittance is:

Therefore, it is possible to determine values of RC and XM in the open-circuit test.

Since the input voltage is low, the current flowing through the excitation branch is

negligible; therefore, all the voltage drop in the transformer is due to the series

elements in the circuit. The magnitude of the series impedance referred to the

primary side of the transformer is:


SCSE

SC

VZ

I

cos SC

SC SC

PPF

V I

The short-circuit test:.

Fairly low input voltage is applied to the

primary side of the transformer. This voltage

is adjusted until the current in the secondary

winding equals to its rated value.

The input voltage, current, and power are measured.

The power factor of the current is given by:


0SC SCSE

SC SC

V VZ

I I

2 2

SE eq eq

SE p S p S

Z R jX

Z R a R j X a X

Therefore:

Since the serial impedance ZSE is equal to

The same tests can be performed on the secondary side of the transformer. The

results will yield the equivalent circuit impedances referred to the secondary

side of the transformer.

Example

Example 4.2: We need to determine the equivalent circuit impedances of a 20

kVA, 8000/240 V, 60 Hz transformer. The open-circuit and short-circuit tests led to

the following data:

VOC = 8000 V VSC = 489 V

IOC = 0.214 A ISC = 2.5 A

POC = 400 W PSC = 240 W

1 1159 ; 38.3

0.0000063 0.0000261C MR k X k

38.3 ; 192eq eqR X

The per-unit system

1. Note that the apparent power rating (base) of a transformer is

the same for both the primary and secondary sides.

2. On the other hand, the voltage rating (base) changes

according to its turn ratio.

3. The impedance of the transformer is often given in pu.

Example: a 110/440 V, 2.5 kVA transformer has a leakage

reactance of 1.24% (or 0.0124 pu). Determine the Ohmic

value when referred to the low-voltage side:

Ans. 60 mΩ.

The per-unit system: Example

1 1

2 2

11

1

,

,

,

8 000 ; 20 000

8 0003 200

20 000

38.4 1920.012 0.06

3 200

159 00049.7

3 200

0012

3 200

base base

basebase

base

SE pu

C pu

M pu

V V S VA

VZ

S

jZ j pu

R pu

X pu

Example 4.4: Sketch the appropriate per-unit equivalent circuit for a

8000/240 V, 60 Hz, 20 kVA transformer with Rc = 159 k, XM = 38.4 k, Req

= 38.3 , Xeq = 192 . All the impedance values are referred to the primary

side

To convert the transformer to per-unit system, the primary circuit base

impedance needs to be found.

Voltage Regulation (VR)

, , ,

, ,

100% 100%s nl s fl p s fl

s fl s fl

V V V a VVR

V V

, , ,

, ,

100%p pu s fl pu

s fl pu

V VVR

V

Since a real transformer contains series impedances, the transformer’s

output voltage varies with the load even if the input voltage is constant.

To compare transformers in this respect, the quantity called a full-load

voltage regulation (VR) is defined as follows:

In a per-unit system:

Note: the VR of an ideal transformer is zero under any load condition.

Where Vs,nl and Vs,fl are the secondary no load and full load voltages.

Transformer phasor diagram

p

s eq s eq s

VV R I jX I

a

To determine the VR of a transformer, it is necessary to understand the voltage

drops within it. Usually, the effects of the excitation branch on transformer VR can

be ignored and, therefore, only the series impedances need to be considered. The

VR depends on the magnitude of the impedances and on the current phase angle.

A phasor diagram is often used in the VR determinations. The phasor voltage Vs is

assumed to be at 00 and all other voltages and currents are compared to it.

Considering the diagram and by applying the

Kirchhoff’s voltage law, the primary voltage is:

A transformer phasor diagram is a graphical

representation of this equation.

Transformer phasor diagram

A transformer operating at a lagging power factor:

It is seen that Vp/a > Vs, VR > 0

A transformer operating at

a unity power factor:

It is seen that VR > 0

A transformer operating at a

leading power factor:

If the secondary current is leading,

the secondary voltage can be higher

than the referred primary voltage;

VR < 0.

Considering the transformer equivalent circuit, we notice three types of losses:

Transformer efficiency

100% 100%out out

in out loss

P P

P P P

cosout s s sP V I

cos100%

cos

s s

Cu core s s

V I

P P V I

The efficiency of a transformer is defined as:

Note: the same equation describes the efficiency of motors and generators.

1. Copper (I2R) losses – are accounted for by the series resistance

2. Hysteresis and eddy current losses – are accounted for by the resistor Rc.

Since the output power is

The transformer efficiency is

The transformer efficiency: Example

VOC = 2300 V VSC = 47 V

IOC = 0.21 A ISC = 6.0 A

POC = 50 W PSC = 160 W

Example 4.5: A 15 kVA, 2300/230 V transformer was tested to by open-circuit and

closed-circuit tests. The following data was obtained:

a) Find the equivalent circuit of this transformer referred to the high-voltage side.

b) Find the equivalent circuit of this transformer referred to the low-voltage side.

c) Calculate the full-load voltage regulation at 0.8 lagging power factor, at 1.0

power factor, and at 0.8 leading power factor.

d) Plot the voltage regulation as load is increased from no load to full load at

power factors of 0.8 lagging, 1.0, and 0.8 leading.

e) What is the efficiency of the transformer at full load with a power factor of 0.8

lagging?

The transformer efficiency: Example

100% 98.03%out

Cu core out

P

P P P

Transformer taps and voltage regulation

We assumed before that the transformer turns ratio is a fixed (constant)

for the given transformer. Frequently, distribution transformers have a

series of taps in the windings to permit small changes in their turns ratio.

Typically, transformers may have 4 taps in addition to the nominal setting

with spacing of 2.5 % of full-load voltage. Therefore, adjustments up to 5

% above or below the nominal voltage rating of the transformer are

possible.

Example 4.6: A 500 kVA, 13 200/480 V transformer has four 2.5 % taps

on its primary winding. What are the transformer’s voltage ratios at each

tap setting?

Ans: + 5.0% tap 13 860/480 V

+ 2.5% tap 13 530/480 V

Nominal rating 13 200/480 V

- 2.5% tap 12 870/480 V

- 5.0% tap 12 540/480 V

Transformer taps and voltage regulation

• Taps allow adjustment of the transformer in the field to

accommodate for local voltage variations.

• Sometimes, transformers are used on a power line, whose

voltage varies widely with the load (due to high line impedance,

for instance). Normal loads need fairly constant input voltage

though

• One possible solution to this problem is to use a special

transformer called a tap changing under load (TCUL)

transformer or voltage regulator. TCUL is a transformer with the

ability to change taps while power is connected to it. A voltage

regulator is a TCUL with build-in voltage sensing circuitry that

automatically changes taps to keep the system voltage

constant.

• These “self-adjusting” transformers are very common in

modern power systems.

The autotransformer

Sometimes, it is desirable to change the voltage by a small amount (for

instance, when the consumer is far away from the generator and it is

needed to raise the voltage to compensate for voltage drops).

In such situations, it would be expensive to wind a transformer with two

windings of approximately equal number of turns. An autotransformer (a

transformer with only one winding) is used instead.

Diagrams of step-up and step-down autotransformers:

Common

winding

Series

winding

Series

winding

Common

winding

Output (up) or input (down) voltage is a sum of voltages across common and series windings.

The autotransformer

Since the autotransformer’s coils are physically connected, a different

terminology is used for autotransformers:

The voltage across the common winding is called a common voltage VC,

and the current through this coil is called a common current IC. The voltage

across the series winding is called a series voltage VSE, and the current

through that coil is called a series current ISE.

The voltage and current on the low-voltage side are called VL and IL; the

voltage and current on the high-voltage side are called VH and IH.

For the autotransformers:

C C

SE SE

V N

V N

C C SE SEN I N ICL

H C SE

NV

V N N

C SEL

H C

N NI

I N

The apparent power advantage

The ratio of the apparent power in the primary and secondary of the

autotransformer to the apparent power actually traveling through its

windings is

IO SE C

W SE

S N N

S N

The last equation described the apparent power rating advantage of

an autotransformer over a conventional transformer.

SW is the apparent power actually passing through the windings. The rest

passes from primary to secondary parts without being coupled through the

windings.

Note that the smaller the series winding, the greater the advantage!

The apparent power advantage

For example, a 5 MVA autotransformer that connects a 110 kV system to a

138 kV system would have a turns ratio (common to series) 110:28. Such

an autotransformer would actually have windings rated at:

285 1.015

28 110

SEW IO

SE C

NS S MVA

N N

Therefore, the autotransformer would have windings rated at slightly over 1

MVA instead of 5 MVA, which makes is 5 times smaller and, therefore,

considerably less expensive.

However, the construction of autotransformers is usually slightly different.

In particular, the insulation on the smaller coil (the series winding) of the

autotransformer is made as strong as the insulation on the larger coil to

withstand the full output voltage.

The primary disadvantage of an autotransformer is that there is a direct

physical connection between its primary and secondary circuits. Therefore,

the electrical isolation of two sides is lost.

3-phase transformers

The majority of the power generation/distribution systems in the world are 3-

phase systems. The transformers for such circuits can be constructed either

as a 3-phase bank of independent identical transformers (can be replaced

independently) or as a single transformer wound on a single 3-legged core

(lighter, cheaper, more efficient).

3-phase transformer connections

We assume that any single transformer in a 3-phase transformer (bank)

behaves exactly as a single-phase transformer. The impedance, voltage

regulation, efficiency, and other calculations for 3-phase transformers are

done on a per-phase basis, using the techniques studied previously for

single-phase transformers.

Four possible connections for a 3-phase transformer bank are:

1. Y-Y

2. Y-

3. -

4. -Y


1. Y-Y connection: The primary voltage on each phase of

the transformer is

3

LPP

VV

The secondary phase voltage is

3LS SV V

The overall voltage ratio is

3

3

PLP

LS S

VVa

V V


4. - connection:

The primary voltage on each phase of

the transformer is

P LPV V


LS SV V


PLP

LS S

VVa

V V

No phase shift, no problems with

unbalanced loads or harmonics.


3. -Y connection:

The primary voltage on each phase of

the transformer is

P LPV V


3LS SV V


3 3

PLP

LS S

VV a

V V

The same advantages and the same

phase shift as the Y- connection.

Transferring Ohmic Values of Per-Phase Impedances

Phase Shifts and Equivalent Circuits

• When stepping up the voltage in a Y-Δ or Δ-Y transformer, the positive sequence voltages and currents are advanced by 30o.

Per-Unit Impedances in 3-Winding Transformer

Zp, Zs, and Zt, are the impedances of the primary, secondary, and tertiary

windings, respectively, referred to the primary circuit. if Zps, Zpt, and Zst, are

the measured impedances referred to the primary circuit, then

Fictitious

neutral point

Regulating Transformers for Control of Voltage Magnitude and Phase

Control of Voltage Magnitude

Control of Voltage Phase Angle

Example (refer to book)

Case 0: Two parallel transformers with the same

turn ratio and same rating. → the transformers

share the load equally

Case 1: Tb is providing a voltage ratio 5% higher

than Ta. → The transformer with the higher tap

setting is supplying most of the reactive power to

the load.

Case 2: Tb is a regulating transformer with a

phase shift of 3o. → The phase shifting transformer

is supplying most of the real power to the load.

Transformers Operating in Parallel

• Example: a 100 kVA transformer (with X= 4%) is connected in parallel with an existing 250 kVA transformer (with X = 6%). The transformers are rated at 7200V/240V, and they supply a load of 330 kVA at nominal voltage. Calculate the following:

1. The nominal secondary current of each transformer: (Ans. 416.7 A and 1,041A)

2. The current and impedance of the load (Ans. 1,375 A and 174.5 mΩ)

3. The impedance of each transformer when referred to the secondary side: (Ans. 23 mΩ and 13.8 mΩ)

4. The current flow through each transformer secondary winding (Ans. 515.6A and 859.4)

5. Which transformer, if any, is overloaded? (Ans. The smaller transformer is carrying 124 kVA – nearly a 25% overload!)

Transformer ratings: Voltage and Frequency

The voltage rating is a) used to protect the winding insulation from breakdown;

b) related to the magnetization current of the transformer (more important)

If a steady-state voltage

( ) sinMv t V t

is applied to the transformer’s

primary winding, the transformer’s

flux will be

( ) c) o1

s( M

p p

v t dV

t tN

tN

An increase in voltage will lead to a

proportional increase in flux.

However, after some point (in a

saturation region), such increase in

flux would require an unacceptable

increase in magnetization current!

flux

Magnetization

current

Transformer ratings: Voltage and Frequency

Therefore, the maximum applied voltage (and thus the rated voltage) is set by

the maximum acceptable magnetization current in the core.

We notice that the maximum flux is also related to the frequency:

maxmax

p

V

N

Therefore, to maintain the same maximum flux, a change in frequency (say, 50

Hz instead of 60 Hz) must be accompanied by the corresponding correction in

the maximum allowed voltage. This reduction in applied voltage with frequency

is called derating. As a result, a 50 Hz transformer may be operated at a 20%

higher voltage on 60 Hz if this would not cause insulation damage.

Transformer ratings: Apparent Power

The apparent power rating sets (together with the voltage rating) the

current through the windings. The current determines the i2R losses and,

therefore, the heating of the coils.

Overheating shortens the life of transformer’s insulation. Under any

circumstances, the temperature of the windings must be limited.

A typical power transformer rating would be OA/FA/FOA where

• The OA (open-air) rating is the base power rating.

• The FA (forced-air) rating is the power carrying capability when

forced air cooling (fans) are used.

• The FOA (forced-oil-air) rating is the power carrying capability when

forced air cooling (fans) are used in addition to oil circulating pumps.

Each cooling method provides nearly an additional 1/3 capability. For

example, a common substation transformer is rated at 21/28/35 MVA.

Transformer ratings: Current inrush

Assuming that the following voltage is applied to the transformer at the moment

it is connected to the line:

( ) sinMv t V t

The maximum flux reached on the first half-cycle depends on the phase of the

voltage at the instant the voltage is applied. If the initial voltage is

( ) sin 90 cosM Mv t V t V t

and the initial flux in the core is zero, the maximum flux during the first half-cycle

is equals to the maximum steady-state flux (which is ok):

maxM

p

V

N

However, if the voltage’s initial phase is zero, i.e.

( ) sinMv t V t

Transformer ratings: Current inrush

the maximum flux during the first half-cycle will be

0

max

0

1si

2n cos MM

M

p pp

VV t dt t

N N

V

N

Which is twice higher than a normal steady-state flux!

Doubling the maximum flux in the core

can bring the core in a saturation and,

therefore, may result in a huge

magnetization current!

Normally, the voltage phase angle cannot

be controlled. As a result, a large inrush

current is possible during the first several

cycles after the transformer is turned ON.

The transformer and the power system

must be able to handle these currents.

Practice Problems (Chap. 2)

6, 8, 13, 15, 19, 21, 23.

ee 340 power transformers -...

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